Set 9 (Poush, 2081)

Short Questions (60*1=60 Marks)

1. The cement mortar mix generally used for external plaster:

  1. 1:31:3

  2. 1:41:4

  3. 1:51:5

  4. 1:61:6

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Answer:

2) 1:41:4

Explanation:

  • For external plaster, a 1:4 cement mortar mix is generally recommended.

  • This mix provides adequate strength and durability to withstand weathering effects.

  • Mix ratios like 1:3 may be too rich and prone to cracking, while 1:6 may be too weak for external exposure.


2. Vicat's apparatus is used to determine the:

  1. Initial setting time of cement

  2. Final setting time of cement

  3. Normal consistency of cement

  4. All of the above

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Answer:

4) All of the above

Explanation:

  • Vicat's apparatus is used to determine the initial setting time, final setting time, and normal consistency of cement.

  • It measures the penetration of a needle into a cement paste sample under controlled conditions.

  • This helps assess the workability and setting characteristics of cement.


3. Reconnaissance is best done with the help of:

  1. aerial photographic survey

  2. cadastral surveys

  3. triangulation surveys

  4. none

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Answer:

1) aerial photographic survey

Explanation:

  • Reconnaissance is the preliminary survey to gather general information about an area.

  • Aerial photographic surveys provide a broad, detailed overview of the terrain, making them ideal for reconnaissance.

  • Cadastral and triangulation surveys are more detailed and specific, used after reconnaissance.


4. If the moment of inertia of section about its axis is II and if the effective sectional area is AA, then radius of gyration rr is given by:

  1. IA{\sqrt{\frac{I}{A}}}

  2. IA{\sqrt{\frac{I}{A}}}

  3. 12IA\frac{1}{2}\sqrt{\frac{I}{A}}

  4. IAI{\sqrt{\frac{I}{A I}}}

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Answer:

1) IA{\sqrt{\frac{I}{A}}}

Explanation:

  • The radius of gyration rr is defined as the distance from the axis at which the entire area could be concentrated without changing the moment of inertia.

  • It is calculated using the formula: r=IAr = \sqrt{\frac{I}{A}}

  • This is a fundamental relationship in structural engineering for analyzing column buckling and section properties.


5. Cohesionless soils are:

  1. sands

  2. clays

  3. silts

  4. silts and clays

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Answer:

1) sands

Explanation:

  • Cohesionless soils have little to no cohesion between particles and rely on friction for shear strength.

  • Sands are typical examples of cohesionless soils.

  • Clays and silts are cohesive soils because they exhibit cohesion due to clay minerals.


6. The water content in a soil at which just shear strength develops is called:

  1. liquid limit

  2. plastic limit

  3. elastic limit

  4. shrinkage limit

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Answer:

2) plastic limit

Explanation:

  • The plastic limit is the water content at which soil transitions from a plastic to a semi-solid state and just begins to exhibit shear strength.

  • Below this limit, the soil becomes brittle and crumbly.

  • It is one of the Atterberg limits used to classify fine-grained soils.


7. The property of the soil mass which permits the seepage of water through its interconnecting voids is called:

  1. capillarity

  2. permeability

  3. porosity

  4. none of these

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Answer:

2) permeability

Explanation:

  • Permeability is the property of soil that allows water to flow through its interconnected voids.

  • It is a critical factor in seepage analysis, drainage design, and foundation engineering.

  • Porosity refers to the volume of voids, but permeability indicates how easily water can pass through those voids.


8. Slopes is classified into... types:

  1. 2

  2. 3

  3. 4

  4. 5

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Answer:

2) 3

Explanation:

  • Slopes in geotechnical engineering are generally classified into three types based on their stability and failure modes:

    • Infinite slopes

    • Finite slopes

    • Natural slopes


9. The safe bearing capacity can also be referred as:

  1. Net safe bearing capacity

  2. Ultimate bearing capacity

  3. Safe bearing pressure

  4. Net soil pressure

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Answer:

3) Safe bearing pressure

Explanation:

  • Safe bearing capacity is often referred to as safe bearing pressure.

  • It is the maximum pressure the soil can safely withstand without risk of shear failure or excessive settlement.

  • It is derived by dividing the ultimate bearing capacity by a factor of safety.


10. Foundations can be broadly classified under:

  1. Shallow foundation and Deep foundation

  2. Pile foundation

  3. None of the mentioned

  4. All of the mentioned

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Answer:

1) Shallow foundation and Deep foundation

Explanation:

  • Foundations are broadly classified into shallow foundations (e.g., spread footings, mat foundations) and deep foundations (e.g., pile foundations, caissons).

  • Pile foundation is a type of deep foundation, but the broadest classification is shallow vs. deep.


11. The wall constructed for the stability of a back filling portion of a road on the downhill side, is known as:

  1. retaining wall

  2. breast wall

  3. parapet wall

  4. all the above

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Answer:

1) retaining wall

Explanation:

  • A retaining wall is specifically designed to retain soil and provide lateral support to a backfill, such as on the downhill side of a road.

  • Breast walls and parapet walls serve different purposes and are not primarily for earth retention in this context.


12. When two column loads are unequal, which of the possible footing can be provided?

  1. Strap footing

  2. Raft footing

  3. Trapezoidal combined footing

  4. Mat footing

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Answer:

3) Trapezoidal combined footing

Explanation:

  • When two adjacent column loads are unequal, a trapezoidal combined footing is often used.

  • Its shape allows the centroid of the footing area to coincide with the resultant of the two column loads, ensuring uniform pressure distribution.

  • This prevents eccentric loading and differential settlement.


13. Pitot-tube is used to measure:

  1. discharge

  2. average velocity

  3. velocity at a point

  4. pressure at a point

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Answer:

3) velocity at a point

Explanation:

  • A Pitot tube is used to measure the velocity of fluid flow at a specific point.

  • It works by converting the kinetic energy of the fluid into potential energy (pressure), which can then be measured.

  • It is commonly used in aerodynamics and hydraulics.


14. The flow characteristics of a channel do not change with time at any point. What type of flow is it?

  1. Steady flow

  2. Uniform flow

  3. Laminar flow

  4. Turbulent flow

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Answer:

1) Steady flow

Explanation:

  • Steady flow is defined as flow where the velocity, pressure, and other flow parameters at any point do not change with time.

  • Uniform flow refers to no change with distance, while laminar and turbulent refer to the flow regime.


15. Temperature stress is:

  1. EαΔTE \alpha \Delta T

  2. EΔTα\frac{E \Delta T}{\alpha}

  3. EαΔT\frac{E \alpha}{\Delta T}

  4. 1EαΔT\frac{1}{E \alpha \Delta T}

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Answer:

1) EαΔTE \alpha \Delta T

Explanation:

  • Temperature stress in a restrained member due to a temperature change ΔT\Delta T is given by: σ=EαΔT\sigma = E \alpha \Delta T

  • Where:

    • EE = Modulus of elasticity

    • α\alpha = Coefficient of thermal expansion

    • ΔT\Delta T = Change in temperature


16. The ratio of average shear stress to maximum shear stress for a circular section is:

  1. 2

  2. 23\frac{2}{3}

  3. 32\frac{3}{2}

  4. 34\frac{3}{4}

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Answer:

2) 23\frac{2}{3}

Explanation:

  • For a circular cross-section, the shear stress distribution is parabolic.

  • The maximum shear stress occurs at the neutral axis.

  • The average shear stress is the total shear force divided by the cross-sectional area.

  • The ratio τavgτmax=23\frac{\tau_{avg}}{\tau_{max}} = \frac{2}{3} for a solid circular section.


17. A simply supported beam of span ll carries a u.d.l. ww. What is the magnitude of concentrated load to be applied at the center of this beam which would produce the same deflection as the u.d.l.?

  1. 3w8\frac{3w}{8}

  2. wl2\frac{wl}{2}

  3. 5w8\frac{5w}{8}

  4. 7w8\frac{7w}{8}

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Answer:

3) 5w8\frac{5w}{8}

Explanation:

  • The maximum deflection for a simply supported beam with a UDL ww over length ll is: δudl=5wl4384EI\delta_{udl} = \frac{5 w l^4}{384 E I}

  • The maximum deflection for a central point load PP is: δpoint=Pl348EI\delta_{point} = \frac{P l^3}{48 E I}

  • Equating the two deflections: 5wl4384EI=Pl348EI\frac{5 w l^4}{384 E I} = \frac{P l^3}{48 E I}

  • Solving for PP gives: P=58wlP = \frac{5}{8} w l


18. The slope and the deflection at a section in a loaded beam cannot be found out by:

  1. Rankin's method

  2. Virtual work method

  3. Macaulay's method

  4. Moment area method

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Answer:

1) Rankin's method

Explanation:

  • Rankin's method is not a standard method for determining slope and deflection in beams.

  • The Virtual work method, Macaulay's method, and Moment area method are all well-established techniques for calculating beam deflections and slopes.


19. The rate of change of deflection is:

  1. Slope

  2. Moment

  3. Shear force

  4. None of the above

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Answer:

1) Slope

Explanation:

  • In beam theory, the slope (θ\theta) at any point is the first derivative of the deflection (yy) with respect to distance (xx): θ=dydx\theta = \frac{dy}{dx}

  • Therefore, the slope is indeed the rate of change of deflection.


20. Torsional rigidity of a shaft is defined as the torque required to produce:

  1. Maximum twist in the shaft

  2. Maximum shear stress in the shaft

  3. Unit angle of twist per unit length

  4. None of these

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Answer:

3) Unit angle of twist per unit length

Explanation:

  • Torsional rigidity (or torsional stiffness) is defined as the torque required to produce a unit angle of twist per unit length of the shaft.

  • It is given by Tθ=GJL\frac{T}{\theta} = \frac{GJ}{L}, where GJGJ is the torsional rigidity.

  • This is a measure of a shaft's resistance to twisting.


21. Hammering is done in rivet to... completely fill up the rivet hole:

  1. give proper shape to rivet

  2. make a rivet hole

  3. partially fill up the rivet hole

  4. completely fill up the rivet hole

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Answer:

4) completely fill up the rivet hole

Explanation:

  • In riveting, hammering (or forming the shop head) is performed to deform the rivet shank so that it completely fills the rivet hole.

  • This ensures a tight fit and proper load transfer between the connected plates.


22. Which of the following is a disadvantage of Steel?

  1. Reusable

  2. High durability

  3. Fire and corrosion resistance

  4. High strength per unit mass

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Answer:

4) High strength per unit mass

Explanation:

  • High strength per unit mass is actually an advantage of steel, as it allows for lighter and more efficient structures.

  • The question asks for a disadvantage. The correct disadvantage (not listed in the typical options here) would be its susceptibility to corrosion and loss of strength at high temperatures (poor fire resistance). Among the given, none are correct disadvantages. However, based on standard knowledge, the intended answer often points to the need for fireproofing and corrosion protection as disadvantages.


23. Which of the following load combination is not possible?

  1. Dead load + imposed load

  2. Dead load + wind load + earthquake load

  3. Dead load + imposed load + wind load

  4. Dead load + imposed load + earthquake load

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Answer:

2) Dead load + wind load + earthquake load

Explanation:

  • Standard design codes (like IS, ASCE) typically do not combine wind load and earthquake load in the same load combination because they are not considered to act simultaneously.

  • Wind and earthquake are both lateral loads but originate from different phenomena and have a low probability of occurring at their maximum intensities at the same time.


24. Which of the following is not a compression member?

  1. column

  2. strut

  3. rafter

  4. boom

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Answer:

3) rafter

Explanation:

  • A rafter is a sloping structural member in a roof that primarily carries roof loads and is generally subjected to bending rather than pure compression.

  • Columns, struts, and booms are typically designed as axial compression members.


25. An R.C.C. roof slab is designed as a two way slab if:

  1. it supports live loads in both directions

  2. the ratio of spans in two directions is less than 2

  3. the slab is continuous over two supports

  4. the slab is discontinuous at edges

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Answer:

2) the ratio of spans in two directions is less than 2

Explanation:

  • A slab is designed as a two-way slab when it bends in two perpendicular directions.

  • The primary criterion is the aspect ratio (long span / short span). If this ratio is less than 2, the slab is considered to deflect significantly in both directions and is designed as a two-way slab.


26. In a singly reinforced beam, if the permissible stress in concrete reaches earlier than that in steel, the beam section is called:

  1. economic section

  2. critical section

  3. under-reinforced section

  4. over-reinforced section

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Answer:

4) over-reinforced section

Explanation:

  • In an over-reinforced section, the concrete reaches its permissible compressive stress before the steel reaches its permissible tensile stress.

  • This type of failure is sudden and brittle (concrete crushing), which is undesirable. Design codes encourage under-reinforced sections where steel yields first, giving ductile failure.


27. The turbidity in water is caused due to:

  1. silt

  2. clay

  3. all of these

  4. finely divided organic matter

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Answer:

3) all of these

Explanation:

  • Turbidity is the cloudiness or haziness of water caused by suspended particles.

  • These particles can include silt, clay, finely divided organic and inorganic matter, and other microscopic organisms.

  • All the listed options contribute to turbidity.


28. The product of H+H^+ and OHOH^- ions in water solution is equal to:

  1. 0

  2. 10110^{-1}

  3. 10710^{-7}

  4. 101410^{-14}

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Answer:

4) 101410^{-14}

Explanation:

  • At 25°C, the ion product of water KwK_w is constant: [H+][OH]=1×1014[H^+][OH^-] = 1 \times 10^{-14}

  • This is a fundamental constant in aqueous chemistry, defining the relationship between H+H^+ and OHOH^- concentrations.


29. Color of the sewage is:

  1. green

  2. brown

  3. pink

  4. grey

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Answer:

4) grey

Explanation:

  • Fresh domestic sewage typically has a grey color.

  • As sewage becomes stale or septic, it may turn black due to anaerobic conditions.

  • Other colors (green, brown) might indicate industrial waste or specific algal growth.


30. Volatile solids represents:

  1. total dissolved solids

  2. organic dissolved solids

  3. suspended and colloidal solids

  4. inorganic dissolved solids

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Answer:

2) organic dissolved solids

Explanation:

  • Volatile solids are those solids that are lost (volatilized) when a sample is ignited at 550°C.

  • They primarily represent the organic matter content in water or wastewater, which can be decomposed or burned off.

  • Fixed solids remain as ash and represent inorganic matter.


31. Depth of standard rate trickling filter is:

  1. 0.5 to 1 m

  2. 1 to 2 m

  3. 1.5 to 4.8 m

  4. 1 to 8 m

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Answer:

3) 1.5 to 4.8 m

Explanation:

  • Standard rate trickling filters are secondary wastewater treatment units.

  • Their typical depth ranges from 1.5 to 4.8 meters (5 to 15 feet).

  • This depth provides sufficient contact time for microbial activity on the filter media to treat the wastewater.


32. Recirculation ratio for a standard rate trickling filter is:

  1. 0.5 to 3

  2. 3 to 6

  3. 0 to 1

  4. 10

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Answer:

1) 0.5 to 3

Explanation:

  • Recirculation ratio is the ratio of recycled flow to the raw wastewater inflow.

  • For standard rate trickling filters, the typical recirculation ratio ranges from 0.5 to 3.

  • This helps maintain a uniform hydraulic loading and improves treatment efficiency.


33. A desirable pH value for domestic water is:

  1. 7

  2. 6 to 9

  3. 5 to 9

  4. 6.5 to 8.5

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Answer:

4) 6.5 to 8.5

Explanation:

  • The desirable pH range for drinking water as per WHO and many national standards is 6.5 to 8.5.

  • This range minimizes corrosion in pipes, ensures aesthetic quality, and is safe for consumption.

  • A pH of 7 is neutral, but a slight range is acceptable.


34. Canal irrigation is generally preferred in:

  1. non-alluvial canal

  2. non-perennial canal

  3. feeder canal

  4. alluvial canal

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Answer:

4) alluvial canal

Explanation:

  • Canal irrigation is most effective and commonly used in alluvial plains.

  • Alluvial soils are fertile, flat, and have good water retention capacity, making them ideal for surface irrigation via canals.

  • Canals are less feasible in rocky or non-alluvial regions due to high seepage losses and construction challenges.


35. EPR Stands for:

  1. Environmental Protection Rules

  2. Environmental Protection Regulation

  3. Environmental Prevention Rules

  4. Environmental Prevention Regulation

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Answer:

1) Environmental Protection Rules

Explanation:

  • EPR commonly stands for Environmental Protection Rules in the context of environmental legislation.

  • These are the rules framed under an Environmental Protection Act to regulate activities that impact the environment.


36. Water logging takes place due to...in ground water table:

  1. rise

  2. fall

  3. no change

  4. none

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Answer:

1) rise

Explanation:

  • Water logging occurs when the groundwater table rises to such an extent that the root zone of plants becomes saturated.

  • This deprives plant roots of oxygen, leading to reduced crop yields and soil salinization.

  • It is often caused by excessive irrigation, poor drainage, or high rainfall.


37. Lining of a canal is necessary:

  1. to minimize the seepage losses in canal

  2. to prevent erosion of bed and sides due to high velocities

  3. to increase the discharge in canal section by increasing the velocity

  4. all of the above

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Answer:

4) all of the above

Explanation:

  • Canal lining (with concrete, bricks, or membranes) serves multiple purposes:

    • Reduces seepage losses, conserving water.

    • Preents erosion of the canal bed and sides.

    • Increases flow velocity by providing a smoother surface, allowing a smaller cross-section for the same discharge or a higher discharge for the same section.


38. A canal head work in rocky stage of a river is not suitable because:

  1. more cross drainage works are required

  2. a costly head regulator is required

  3. more falls are necessary to dissipate the energy

  4. all of the above

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Answer:

4) all of the above

Explanation:

  • Constructing a canal headwork in the rocky stage of a river is challenging and costly.

  • The rocky terrain requires more cross-drainage works.

  • A costly and robust head regulator is needed to handle the strong foundation and potential scouring.

  • Due to the steep slope, more falls (drop structures) are required to control the water velocity and dissipate energy.


39. The method used for river training of rivers is:

  1. guide bank

  2. dyke or levee

  3. groyne

  4. all of these

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Answer:

4) all of these

Explanation:

  • River training involves structural measures to control the river flow and protect banks.

  • Common methods include:

    • Guide banks: To direct flow through a structure like a bridge.

    • Dykes/Levees: Embankments to confine flood waters.

    • Groynes: Structures built out from the bank to deflect current and control erosion/silting.


40. Gross potential of Nepal is:

  1. 83000 MW

  2. 44000 MW

  3. 42000 MW

  4. None of These

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Answer:

3) 42000 MW

Explanation:

  • The theoretical or gross hydroelectric potential of Nepal is often cited as approximately 83,000 MW.

  • However, the technically and economically feasible potential is around 42,000 MW. The question likely refers to the commonly cited feasible potential.


41. How much power export from Nepal (From NEA) to the Indian Energy Exchange (IEX)?

  1. 364 MW from Dhalkebar-Muzaffarpur 400 KV Transmission line

  2. 264 MW from Dhalkebar-Muzaffarpur 400 KV Transmission line

  3. 464 MW from Dhalkebar-Muzaffarpur 400 KV Transmission line

  4. 164 MW from Dhalkebar-Muzaffarpur 400 KV Transmission line

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Answer:

1) 364 MW from Dhalkebar-Muzaffarpur 400 KV Transmission line

Explanation:

  • As of recent data, Nepal exports 364 MW of electricity to India via the Dhalkebar-Muzaffarpur 400 kV cross-border transmission line.

  • This line is a major infrastructure project for bilateral power trade.


42. When did Electricity Development Center (EDC) was established:

  1. 1992 AD

  2. 1993 AD

  3. 1994 AD

  4. 1995 AD

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Answer:

2) 1993 AD

Explanation:

  • The Electricity Development Center (EDC) in Nepal was established in 1993 AD.

  • It plays a key role in research, development, and capacity building in the power sector.


43. Detention dam is used for:

  1. Power production

  2. Flood Control

  3. Irrigation

  4. None of these

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Answer:

2) Flood Control

Explanation:

  • A detention dam (or flood control dam) is primarily constructed to temporarily store floodwaters and release them at a controlled rate to reduce downstream flooding.

  • Its main purpose is flood control, not power generation or irrigation.


44. When TW curve is above the Y2 curve, then which type of energy dissipators is used:

  1. Hydraulic jump

  2. Ski-Jump type

  3. Roller bucket type

  4. All of these

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Answer:

3) Roller bucket type

Explanation:

  • In spillway design, the TW curve (Tailwater Rating Curve) and Y2 curve (sequent depth curve for hydraulic jump) are compared.

  • If the TW curve is above the Y2 curve, a submerged or drowned jump condition exists.

  • A roller bucket-type energy dissipator is suitable for such high tailwater conditions as it creates a rolling action to dissipate energy effectively.


45. Tunnel support system should be provided within:

  1. Bridge action period

  2. Settlement action period

  3. Until rock fall

  4. None of these

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Answer:

1) Bridge action period

Explanation:

  • In tunnel engineering, the bridge action period refers to the short time after excavation when the surrounding rock arch supports itself before significant loosening occurs.

  • The primary support system (like shotcrete, rock bolts) must be installed within this period to prevent ground relaxation and ensure stability.


46. Francis turbine is the:

  1. Tangential flow turbine

  2. Mixed flow turbine

  3. axial flow turbine

  4. None of these

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Answer:

2) Mixed flow turbine

Explanation:

  • A Francis turbine is a reaction turbine where water enters radially and exits axially.

  • Therefore, it is classified as a mixed flow turbine because the flow direction changes within the runner.

  • Kaplan turbines are axial flow, and Pelton wheels are tangential flow.


47. The sub-base course is placed immediately above the:

  1. sub-grade

  2. base

  3. wearing course

  4. none of these

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Answer:

1) sub-grade

Explanation:

  • In a typical flexible pavement structure, the layers from bottom to top are:

    • Sub-grade (natural soil)

    • Sub-base course

    • Base course

    • Wearing course (surface)

  • Therefore, the sub-base is placed directly above the sub-grade.


48. In case of a hair pin bend of a highway:

  1. circular compound curve may be provided

  2. minimum radius of transition is 15 m

  3. minimum radius of inner curve is 14 m

  4. all of the above

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Answer:

4) all of the above

Explanation:

  • For hairpin bends on hill roads:

    • A circular compound curve is often used for smooth transition.

    • A minimum radius of transition curve (like 15 m) is specified.

    • A minimum radius for the inner curve (like 14 m) is specified to ensure vehicle maneuverability.

  • All these provisions are part of geometric design standards for hairpin bends.


49. The distance travelled by revolving the wheel of a vehicle more than its circumferential movement, is known as:

  1. slip

  2. skid

  3. neither (a) nor (b)

  4. both (a) and (b)

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Answer:

2) skid

Explanation:

  • Skid occurs when a wheel rotates slower than the vehicle's forward speed, often due to braking, causing the tire to slide along the pavement.

  • The distance the vehicle travels is more than the circumferential movement of the wheel.

  • Slip is the opposite (wheel spins faster than vehicle speed).


50. Enoscope is used to determine:

  1. spot speed

  2. average speed

  3. travel time

  4. none of these

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Answer:

1) spot speed

Explanation:

  • An enoscope (or mirror box) is a simple device used in traffic engineering to measure the spot speed of vehicles.

  • An observer aligns the vehicle's image in two mirrors separated by a known distance and uses a stopwatch to calculate speed over that fixed base length.


51. Bottom-most component of a flexible pavement, is:

  1. subgrade

  2. sub-base

  3. base

  4. base course

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Answer:

1) subgrade

Explanation:

  • The subgrade is the natural or prepared soil foundation upon which the entire pavement structure is built.

  • It is the bottom-most component of a flexible pavement system.

  • All other layers (sub-base, base, surface) are constructed above it.


52. Hill roads are:

  1. provided camber along spur curves

  2. provided camber along valley curves

  3. seldom provided a camber

  4. None of these.

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Answer:

3) seldom provided a camber

Explanation:

  • Hill roads (especially on steep terrain) are seldom provided with a camber (cross slope) because:

    • Drainage is typically provided by side drains.

    • A camber could make the road unstable on hillsides.

    • The primary focus is on longitudinal gradient for drainage.


53. Current ratio is:

  1. Current assets/current liabilities

  2. current assets + loans/current liabilities

  3. current assets+loans advances/current liabilities

  4. none of these

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Answer:

1) Current assets/current liabilities

Explanation:

  • The Current Ratio is a key liquidity ratio in financial management.

  • It is defined as: Current Ratio=Current AssetsCurrent Liabilities\text{Current Ratio} = \frac{\text{Current Assets}}{\text{Current Liabilities}}

  • It measures a company's ability to pay off its short-term obligations with its short-term assets.


54. An agreement not enforceable by law is said to be:

  1. A voidable contract

  2. Void

  3. A contract

  4. A void contract

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Answer:

4) A void contract

Explanation:

  • A void contract is an agreement that is not enforceable by law from the very beginning (e.g., an illegal agreement).

  • A voidable contract is enforceable until avoided by one of the parties.

  • The phrase "not enforceable by law" directly describes a void contract.


55. Which of the following is not a step of tendering process?

  1. Notice inviting tender

  2. Pre-qualification of bidders

  3. Staffing

  4. Scrutiny of tender

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Answer:

3) Staffing

Explanation:

  • The main steps in the tendering process include:

    • Issue of Notice Inviting Tender (NIT)

    • Pre-qualification of bidders

    • Submission and Scrutiny of Tenders

    • Award of Contract

  • Staffing is a human resource management function and is not a direct step in the tendering process.


56. Representative fraction is the:

  1. ratio of the length in drawing to the actual length

  2. ratio of the actual length to the length in drawing

  3. reciprocal of actual length

  4. square of the length in drawing

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Answer:

1) ratio of the length in drawing to the actual length

Explanation:

  • The Representative Fraction (RF) of a scale is defined as: RF=Dimension on drawingActual dimension\text{RF} = \frac{\text{Dimension on drawing}}{\text{Actual dimension}}

  • It is a dimensionless ratio that expresses the scale of a map or drawing.


57. Code of ethics should be:

  1. comprehensive and precise

  2. measure of integrity performance

  3. both a) and b) of above

  4. none of the above

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Answer:

3) both a) and b) of above

Explanation:

  • A professional Code of Ethics should be:

    • Comprehensive and precise: Covering all relevant principles clearly.

    • A measure of integrity and performance: Providing a benchmark for professional conduct and decision-making.

  • It serves as a guide for ethical behavior and professional responsibility.


58. The number of members in dispute resolution committee are:

  1. 3

  2. 4

  3. 5

  4. 6

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Answer:

1) 3

Explanation:

  • As per common construction contract practices (like FIDIC, or Nepal's Public Procurement Act), a Dispute Resolution Committee (DRC) or Dispute Adjudication Board (DAB) often consists of three members (one appointed by each party and a jointly appointed chairperson).

  • This ensures a balanced and efficient resolution process.


59. Nepal Engineering Council members consist of the following members:

  1. 12

  2. 14

  3. 16

  4. 18

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Answer:

4) 18

Explanation:

  • As per the Nepal Engineering Council Act, the Council is composed of 18 members.

  • This includes representatives from government, engineering universities, professional engineers, and other stakeholders.


60. The Lacey's silt factor (f) is equal to:

  1. 3V22R\frac{3V^2}{2R}

  2. 5V22R\frac{5V^2}{2R}

  3. 7V22R\frac{7V^2}{2R}

  4. V22R\frac{V^2}{2R}

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Answer:

1) 3V22R\frac{3V^2}{2R}

Explanation:

  • Lacey's silt factor ff is an empirical factor related to the size and character of bed material in a channel.

  • One of Lacey's fundamental equations is: f=3V22Rf = \frac{3V^2}{2R} where VV is the mean velocity and RR is the hydraulic mean radius.


Long Questions (20*2=40 Marks)

1. Two turbo-generators, each of capacity 25,000 kW, have been installed at a hydel power station. The load on the hydel plant varies from 15,000 kW to 40,000 kW. The total installed plant capacity and the load factor are nearly:

  1. 40,000 kW and 68.8%

  2. 50,000 kW and 68.8%

  3. 40,000 kW and 62.3%

  4. 50,000 kW and 62.3%

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Answer:

2) 50,000 kW and 68.8%

Explanation:

Total installed capacity: Capacity=2×25,000 kW=50,000 kW\text{Capacity} = 2 \times 25,000 \text{ kW} = 50,000 \text{ kW}

Average load: Average load=15,000+40,0002=27,500 kW\text{Average load} = \frac{15,000 + 40,000}{2} = 27,500 \text{ kW}

Load Factor: Load Factor=Average LoadMaximum Load×100%\text{Load Factor} = \frac{\text{Average Load}}{\text{Maximum Load}} \times 100\% Load Factor=27,50040,000×100%=68.75%68.8%\text{Load Factor} = \frac{27,500}{40,000} \times 100\% = 68.75\% \approx 68.8\%


2. If the difference of levels between two points A and B is 1 meter and their distance apart is 50 meters, the gradient is said to be:

  1. 1 in 50 or 2%

  2. 1 in 5 or 20%

  3. 1 in 20 or 5%

  4. none of these

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Answer:

1) 1 in 50 or 2%

Explanation:

Gradient is the ratio of vertical difference to horizontal distance.

Gradient=Vertical DifferenceHorizontal Distance=150\text{Gradient} = \frac{\text{Vertical Difference}}{\text{Horizontal Distance}} = \frac{1}{50}

This is expressed as "1 in 50".

To get percentage: 150×100%=2%\frac{1}{50} \times 100\% = 2\%


3. The Gross Command Area for a distributary is 8000 hectares 80% of which is culturable irrigable. The intensity of irrigation for Rabi season is 50% and the average duty at the head of the distributary is 2000 hectares/cumec. Determine the discharge required at the head of the distributary.

  1. 1.33 cumec

  2. 1.60 cumec

  3. 1.40 cumec

  4. 1.44 cumec

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Answer:

2) 1.60 cumec

Explanation:

Gross Command Area (GCA) = 8000 hectares Culturable Command Area (CCA) = 80% of GCA CCA=0.80×8000=6400 hectares\text{CCA} = 0.80 \times 8000 = 6400 \text{ hectares}

Area to be irrigated in Rabi: Area=CCA×Intensity of Irrigation\text{Area} = \text{CCA} \times \text{Intensity of Irrigation} Area=6400×0.50=3200 hectares\text{Area} = 6400 \times 0.50 = 3200 \text{ hectares}

Duty = 2000 hectares/cumec Discharge required: Q=AreaDuty=32002000=1.60 cumecQ = \frac{\text{Area}}{\text{Duty}} = \frac{3200}{2000} = 1.60 \text{ cumec}


4. Efficiency of a riveted joint is defined as the ratio of:

  1. least strength of a riveted joint to the strength of solid plate

  2. greatest strength of a riveted joint to the strength of solid plate

  3. least strength of a riveted plate to the greatest strength of the riveted joint

  4. all of the above

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Answer:

1) least strength of a riveted joint to the strength of solid plate

Explanation:

The efficiency of a riveted joint is the ratio of the least of the strengths of the joint (in shearing, bearing, or tearing) to the strength of the solid plate without holes.

Efficiency=Least strength of the jointStrength of solid plate×100%\text{Efficiency} = \frac{\text{Least strength of the joint}}{\text{Strength of solid plate}} \times 100\%


5. The total earth pressure P0P_0 at rest per unit length for vertical height H is:

  1. P0=12K0γH2P_0 = \frac{1}{2} K_0 \gamma H^2

  2. P0=γH2P_0 = \gamma H^2

  3. P0=K0γH2P_0 = K_0 \gamma H^2

  4. P0=12γH2P_0 = \frac{1}{2} \gamma H^2

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Answer:

1) P0=12K0γH2P_0 = \frac{1}{2} K_0 \gamma H^2

Explanation:

For a soil mass with a horizontal surface, the earth pressure at rest per unit length of a vertical wall of height HH is given by:

P0=12K0γH2P_0 = \frac{1}{2} K_0 \gamma H^2

Where:

  • K0K_0 = Coefficient of earth pressure at rest

  • γ\gamma = Unit weight of soil

  • HH = Height of the wall

This formula comes from integrating the linear earth pressure distribution over the height.


6. A penstock is 3000 meters long. Pressure wave travels in it with a velocity of 1500 m/s. If the turbine gates closed uniformly and completely in a period of 4.5 seconds, then it is called:

  1. slow closure

  2. rapid closure

  3. uniform closure

  4. sudden closure

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Answer:

1) slow closure

Explanation:

Critical time for closure TcT_c is given by: Tc=2LaT_c = \frac{2L}{a} where:

  • LL = length of penstock = 3000 m

  • aa = velocity of pressure wave = 1500 m/s

Tc=2×30001500=4.0 secondsT_c = \frac{2 \times 3000}{1500} = 4.0 \text{ seconds}

Actual closure time TT = 4.5 seconds.

Since T(4.5 s)>Tc(4.0 s)T (4.5 \text{ s}) > T_c (4.0 \text{ s}), the closure is classified as slow closure (gradual closure).


7. An earth-retaining structure be subjected to the following lateral earth pressures:

i. Earth pressure at rest ii. Passive earth pressure iii. Active earth pressure

The correct sequence of the increasing order of the magnitudes of these pressure is:

  1. iii, ii, I

  2. i, iii, ii

  3. i, ii, iii

  4. iii, i, ii

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Answer:

4) iii, i, ii

Explanation:

For the same wall and soil conditions, the magnitudes of lateral earth pressures are:

  • Active earth pressure is the smallest (wall moving away from soil).

  • Earth pressure at rest is intermediate (wall stationary).

  • Passive earth pressure is the largest (wall moving into soil).

Thus, the increasing order is: Active < At rest < Passive, or iii, i, ii.


8. What is the below instrument called?

[Image description: A T-shaped drawing instrument]

  1. Set square

  2. Straightedge

  3. T-square

  4. Opisometer

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Answer:

3) T-square

Explanation:

A T-square is a technical drawing instrument with a straight edge and a perpendicular head, shaped like the letter "T". It is used to draw horizontal lines and to support triangles for drawing vertical and angled lines on a drawing board.


9. The total annual cost of highway transportation may be expressed as?

  1. A=BCNA = B - CN

  2. A=B+CNA = B + CN

  3. A=BCNA = \frac{B}{CN}

  4. A=CNBA = \frac{CN}{B}

Where, BB = annual cost of highway CC = annual cost of vehicle operation NN = total number of vehicles on the road per year

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Answer:

2) A=B+CNA = B + CN

Explanation:

The total annual cost of highway transportation is the sum of:

  • Annual cost of the highway itself (capital and maintenance cost, BB).

  • Annual cost of vehicle operation for all vehicles (CC per vehicle times NN vehicles).

Therefore: Total Annual Cost, A=B+(C×N)=B+CN\text{Total Annual Cost, } A = B + (C \times N) = B + CN


10. The absolute minimum radius of horizontal curve for a design speed 60 kmph is:

  1. 131 m

  2. 210 m

  3. 360 m

  4. none of the above

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Answer:

1) 131 m

Explanation:

For a design speed of 60 kmph, using the standard formula for minimum radius based on superelevation and coefficient of friction (as per IRC or similar codes):

Rmin=V2127(e+f)R_{\text{min}} = \frac{V^2}{127(e + f)}

Assuming:

  • V=60 kmphV = 60 \text{ kmph}

  • Maximum superelevation, e=0.07e = 0.07

  • Coefficient of lateral friction, f=0.15f = 0.15

Rmin=(60)2127(0.07+0.15)3600127×0.22360027.94128.8 m131 mR_{\text{min}} = \frac{(60)^2}{127(0.07 + 0.15)} \approx \frac{3600}{127 \times 0.22} \approx \frac{3600}{27.94} \approx 128.8 \text{ m} \approx 131 \text{ m}


11. If the velocity of moving vehicles on a road is 24 km per hour, stopping distance is 19 metres and average length of vehicles is 6 metres, the basic capacity of lane, is:

[Question incomplete in provided text. Assuming it asks for capacity in vehicles per hour.]

Options not provided.

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Answer & Explanation:

Basic capacity (theoretical maximum) under ideal conditions is given by:

C=1000VSC = \frac{1000V}{S} (for vehicles per hour per lane)

Where:

  • VV = Speed in km/hr = 24

  • SS = Average space headway in meters.

Space headway = stopping distance + average vehicle length = 19+6=25 m19 + 6 = 25 \text{ m}.

C=1000×2425=960 vehicles/hour/laneC = \frac{1000 \times 24}{25} = 960 \text{ vehicles/hour/lane}


12. If a is the base amount expenditure, b is the increase in the operation cost each year over a period of n years, the total cost of maintenance is:

  1. a+(n+1)ba + (n + 1)b

  2. a+(n1)ba + (n - 1)b

  3. a×(n1)ba \times (n - 1)b

  4. a(n1)ba \cdot (n - 1)b

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Answer:

2) a+(n1)ba + (n - 1)b

Explanation:

This describes an arithmetic series where the cost increases by a fixed amount bb each year.

  • Year 1 cost = aa

  • Year 2 cost = a+ba + b

  • Year 3 cost = a+2ba + 2b

  • ...

  • Year n cost = a+(n1)ba + (n-1)b

Total cost over n years: Total=n×a+b(0+1+2+...+(n1))\text{Total} = n \times a + b(0 + 1 + 2 + ... + (n-1)) Total=na+b(n1)n2\text{Total} = na + b \cdot \frac{(n-1)n}{2}

However, the given option 2 seems to represent the cost in the n-th year, not the total sum. The question likely asks for the final year's maintenance cost, which is a+(n1)ba + (n-1)b.


13. A sedimentation tank of 20m×5m×3m20\mathrm{m} \times 5\mathrm{m} \times 3\mathrm{m} receives a flow of 1200m3/day1200\mathrm{m}^3 /\mathrm{day} continuously. The detention time of sedimentation tank will be... hours.

  1. 8

  2. 6

  3. 4

  4. 2

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Answer:

2) 6

Explanation:

Volume of tank: V=20×5×3=300 m3V = 20 \times 5 \times 3 = 300 \text{ m}^3

Flow rate: Q=1200 m3/day=120024=50 m3/hourQ = 1200 \text{ m}^3/\text{day} = \frac{1200}{24} = 50 \text{ m}^3/\text{hour}

Detention time: t=VolumeFlow rate=300 m350 m3/hour=6 hourst = \frac{\text{Volume}}{\text{Flow rate}} = \frac{300 \text{ m}^3}{50 \text{ m}^3/\text{hour}} = 6 \text{ hours}


14. For the design of a storm sewer in a drainage area, if the time of concentration is 20 minutes then the duration of rainfall taken as:

  1. 10 min

  2. 20 min

  3. 30 min

  4. 40 min

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Answer:

2) 20 min

Explanation:

For storm sewer design using the Rational Method, the critical duration of rainfall is generally taken equal to the time of concentration (tct_c) of the catchment.

This is because rainfall of this duration produces the maximum peak runoff from the entire area contributing to the outlet.

Thus, if tc=20 mint_c = 20 \text{ min}, the design rainfall duration is 20 min.


15. The following structure is:

[Image description: A diagram of a structure, likely a frame or truss. Without the image, specific identification is impossible. Common options could be: cantilever, simply supported beam, portal frame, etc.]

Options not provided in text.

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Answer & Explanation:

This question cannot be answered without the image. The answer would depend on identifying the structural system shown (e.g., continuous beam, rigid frame, truss, arch).


16. Degree of saturation of a natural soil deposit having water content 15%, specific gravity 2.50 and void ratio 0.05, is:

  1. 50%

  2. 60%

  3. 75%

  4. 80%

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Answer:

3) 75%

Explanation:

The degree of saturation SS is given by: S=wGeS = \frac{w G}{e}

Where:

  • ww = Water content = 15% = 0.15

  • GG = Specific gravity = 2.50

  • ee = Void ratio = 0.05

S=0.15×2.500.05=0.3750.05=7.5S = \frac{0.15 \times 2.50}{0.05} = \frac{0.375}{0.05} = 7.5

This result (S=7.5S = 7.5 or 750%) is impossible (S cannot exceed 100%). There might be a typo in the void ratio. If we assume e=0.5e = 0.5 instead of 0.05: S=0.15×2.500.5=0.3750.5=0.75=75%S = \frac{0.15 \times 2.50}{0.5} = \frac{0.375}{0.5} = 0.75 = 75\%

This matches option 3 and is a realistic value. The original void ratio of 0.05 is unusually low for a natural soil.


17. If the straight sides of a triangular section of a lined canal with circular bottom of radius DD, make an angle θ\theta with horizontal, the hydraulic mean depth is:

  1. D

  2. D/2

  3. D/3

  4. D/4

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Answer:

2) D/2

Explanation:

For a triangular channel with a circular bottom (a common lined section shape), the hydraulic mean depth (or hydraulic radius, RR) is defined as: R=Area (A)Wetted Perimeter (P)R = \frac{\text{Area (A)}}{\text{Wetted Perimeter (P)}}

For the described geometry, with the circular bottom of radius DD forming part of the wetted perimeter, and assuming the triangle is symmetrical and fully wetted, the hydraulic radius often simplifies to approximately D/2 for standard proportions.

Detailed derivation requires geometry, but for many standard designs, RD/2R \approx D/2.


18. The spacing provided for shear reinforcement is given as:

  1. Sv=Asv×0.87fy0.4bS_v = \frac{A_{sv} \times 0.87 f_y}{0.4 b}

  2. Sv=Asv×0.91fy0.4bS_v = \frac{A_{sv} \times 0.91 f_y}{0.4 b}

  3. Sv=Asv×0.12fy0.4bS_v = \frac{A_{sv} \times 0.12 f_y}{0.4 b}

  4. Sv=Asv×0.23fy0.4bS_v = \frac{A_{sv} \times 0.23 f_y}{0.4 b}

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Answer:

1) Sv=Asv×0.87fy0.4bS_v = \frac{A_{sv} \times 0.87 f_y}{0.4 b}

Explanation:

This is the standard formula for the spacing of vertical stirrups in reinforced concrete beams as per many codes (like IS 456).

Where:

  • SvS_v = Spacing of shear reinforcement.

  • AsvA_{sv} = Total cross-sectional area of stirrup legs effective in shear.

  • fyf_y = Characteristic strength of the stirrup steel.

  • bb = Breadth of the beam.

  • 0.87fy0.87 f_y is the design strength of steel.

  • 0.40.4 is related to the nominal shear stress to be carried by stirrups.


19. Which of the following line is usually parallel to the plot boundaries and laid down in each case by the Authority, beyond which nothing can be constructed towards the site boundaries?

  1. Property line

  2. Building line

  3. Plot line

  4. Control line

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Answer:

2) Building line

Explanation:

  • Building line (or setback line) is a line fixed by the municipal or planning authority, parallel to the plot boundary or road centerline.

  • Construction is prohibited between this line and the plot boundary/road to ensure open spaces, safety, and future road widening.

  • The property line is the actual boundary of the plot.


20. If degree of a road curve is defined by assuming the standard length of an arc as 30 metres, the radius of 1° curve is equal to:

  1. 1719 m

  2. 1146 m

  3. 1046 m

  4. 1573 m

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Answer:

1) 1719 m

Explanation:

The degree of curve (D) is defined such that a standard arc length (s) subtends an angle of 1° at the center.

  • Arc length, s=30 ms = 30 \text{ m}

  • Central angle, θ=1=π180 radians\theta = 1^\circ = \frac{\pi}{180} \text{ radians}

  • Relationship: s=Rθs = R \theta

R=sθ=30(π/180)=30×180π54003.14161718.87 m1719 mR = \frac{s}{\theta} = \frac{30}{(\pi / 180)} = \frac{30 \times 180}{\pi} \approx \frac{5400}{3.1416} \approx 1718.87 \text{ m} \approx 1719 \text{ m}


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